Question

A bullet of mass $m$ moving horizontally with a velocity $v$ strikes a block of wood of mass $M$ and gets embedded in the block. The block is suspended from the ceiling by a massless string. The height to which block rises is

• A. $\frac{v^2}{2g}{\left(\frac{m}{M+m}\right)}^2$
• B. $\frac{v^2}{2g}{\left(\frac{M+m}{m}\right)}^2$
• C. $\frac{v^2}{2g}{\left(\frac{m}{M}\right)}^2$
• D. $\frac{v^2}{2g}{\left(\frac{M}{m}\right)}^2$

Answer 1

Answer: (A)

The situation is as shown in the figure.

Let $V$ be velocity of the block-bullet system just after collision. Then by the law of conservation of linear momentum, we get
$mv=(m+M)V$
$V=\frac{mv}{m+M}$
Let the block rises to a height $h$ .
According to law of conservation of mechanical energy, we get
$\frac{1}{2}\left(m+M\right)V^2=\left(m+M\right)gh$
$h=\frac{V^2}{2g}=\frac{v^2}{2g}{\left(\frac{m}{m+M}\right)}^2$